Answer
The solution is
$ x=3 $ and $ y=4 $.
Work Step by Step
The given system of linear equation is
$ x+2y=11 $
$ x-y=-1 $
The augmented matrix is
$\left[\begin{array}{cc|c}
1& 2 &11 \\
1&-1 &-1 \\
\end{array}\right]$
Perform $R_2\rightarrow R_2- R_1$.
Subtract row one from row two as shown below.
$\left[\begin{array}{cc|c}
1& 2 &11 \\
1-1&-1-2 &-1-11 \\
\end{array}\right]$
Simplify.
$\left[\begin{array}{cc|c}
1& 2 &11 \\
0&-3 &-12 \\
\end{array}\right]$
Perform $R_2\rightarrow \frac{R_2}{-3}$.
Divide row two by $-3 $.
$\left[\begin{array}{cc|c}
1& 2 &11 \\
\frac{0}{-3}&\frac{-3}{-3} &\frac{-12}{-3} \\
\end{array}\right]$
Simplify.
$\left[\begin{array}{cc|c}
1& 2 &11 \\
0&1 &4 \\
\end{array}\right]$
Use back substitution to solve the linear system.
$ x+2y=11 $
$ y=4 $ Substitute into above equation.
$ x+2(4)=11 $
$ x+8=11 $
$ x=11-8 $
$x=3 $.
